# Time until distance gets equal to X between two objects moving in opposite direction

Consider two people moving in opposite direction with speeds U meters/second and V meters/second respectively. The task is to find how long will take to make the distance between them X meters.

Examples:

Input: U = 3, V = 3, X = 3
Output: 0.5
After 0.5 seconds, policeman A will be at distance 1.5 meters
and policeman B will be at distance 1.5 meters in the opposite direction
The distance between the two policemen is 1.5 + 1.5 = 3

Input: U = 5, V = 2, X = 4
Output: 0.571429

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: It can be solved using distance = speed * time. Here, distance would be equal to the given range i.e. distance = X and speed would be the sum of the two speeds because they are moving in the opposite direction i.e. speed = U + V.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the time for which ` `// the two policemen can communicate ` `double` `getTime(``int` `u, ``int` `v, ``int` `x) ` `{ ` `    ``double` `speed = u + v; ` ` `  `    ``// time = distance / speed ` `    ``double` `time` `= x / speed; ` `    ``return` `time``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``double` `u = 3, v = 3, x = 3; ` `    ``cout << getTime(u, v, x); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `    ``// Function to return the time for which ` `    ``// the two policemen can communicate ` `    ``static` `double` `getTime(``int` `u, ``int` `v, ``int` `x)  ` `    ``{ ` `        ``double` `speed = u + v; ` ` `  `        ``// time = distance / speed ` `        ``double` `time = x / speed; ` `        ``return` `time; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `u = ``3``, v = ``3``, x = ``3``; ` `        ``System.out.println(getTime(u, v, x)); ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the time  ` `# for which the two policemen  ` `# can communicate  ` `def` `getTime(u, v, x):  ` ` `  `    ``speed ``=` `u ``+` `v  ` ` `  `    ``# time = distance / speed  ` `    ``time ``=` `x ``/` `speed  ` `    ``return` `time  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``u, v, x ``=` `3``, ``3``, ``3` `    ``print``(getTime(u, v, x))  ` ` `  `# This code is contributed ` `# by Rituraj Jain `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to return the time for which ` `    ``// the two policemen can communicate ` `    ``static` `double` `getTime(``int` `u, ``int` `v, ``int` `x)  ` `    ``{ ` `        ``double` `speed = u + v; ` ` `  `        ``// time = distance / speed ` `        ``double` `time = x / speed; ` `        ``return` `time; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `u = 3, v = 3, x = 3; ` `        ``Console.WriteLine(getTime(u, v, x)); ` `    ``} ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai `

## PHP

 ` `

Output:

```0.5
```

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